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3 years ago

A square has a area of 466.56m squared work out the peremiter

Mathematics

1 answer:

MissTica3 years ago

3 0

Answer:

Perimeter = 86.4 (m)

Step-by-step explanation:

A square (with side x) has an area of 466.56 (m^2)

=> x^2 = 466.56

=> x = sqrt(466.56) = 21.6

=> Perimeter of this square: P = 4(x) = 4(21.6) = 86.4 (m)

Hope this helps!

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Determine the vertex of the function f(x) = 2(x-4)^2+5

spayn [35]

Answer:

(4,5)

Step-by-step explanation:

The vertex is written in vertex form.

y = a(x - b)^2 + c

Vertex: (b,c) Notice the sign change on b.

So for your equation, you get (4,5) No sign change on the 5.

5 0

3 years ago

If tap and bre are supplementary and bre is its own complement, find the measure of tap. explain how you arrived at your answer

Pie

Supplementary angles are those whose sum is equal to 180° while those that are complementary are those whose sum is equal to 90°. Since, we are given that bre is the complement of itself, its measure can be calculated through the equation,
m∠bre + m∠bre = 90°
m∠bre = 45°

Then, for the relationship of bre and tap, we have the equation,
m∠bre + m∠tap = 180°
45° + m∠tap = 180°
m∠tap = 135°

The measure of tap is equal to 135°.

3 0

3 years ago

A race car travels 100 feet in 5 seconds, at this rate of speed, how many feet will the race car travel in a minute

Basile [38]

There are 60 seconds in one minute, and 60/5=12, so the race car would travel 100 feet every 5 seconds 12 times in one minute. You can simply multiply 100 by 12 to get your answer.
100x12=1200
The race car travels 1,200 feet in one minute.

6 0

3 years ago

Read 2 more answers

Four buses carrying 146 high school students arrive to Montreal. The buses carry, respectively, 32, 44, 28, and 42 students. One

Naily [24]

Answer:

The expected value of X is $E(X)=\frac{2754}{73} \approx 37.73$ and the variance of X is $Var(X)=\frac{226192}{5329} \approx 42.45$

The expected value of Y is $E(Y)=\frac{73}{2} \approx 36.5$ and the variance of Y is $Var(Y)=\frac{179}{4} \approx 44.75$

Step-by-step explanation:

(a) Let X be a discrete random variable with set of possible values D and probability mass function p(x). The expected value, denoted by E(X) or $\mu_x$ , is

$E(X)=\sum_{x\in D} x\cdot p(x)$

The probability mass function $p_{X}(x)$ of X is given by

$p_{X}(28)=\frac{28}{146} \\\\p_{X}(32)=\frac{32}{146} \\\\p_{X}(42)=\frac{42}{146} \\\\p_{X}(44)=\frac{44}{146}$

Since the bus driver is equally likely to drive any of the 4 buses, the probability mass function $p_{Y}(x)$ of Y is given by

$p_{Y}(28)=p_{Y}(32)=p_{Y}(42)=p_{Y}(44)=\frac{1}{4}$

The expected value of X is

$E(X)=\sum_{x\in [28,32,42,44]} x\cdot p_{X}(x)$

$E(X)=28\cdot \frac{28}{146}+32\cdot \frac{32}{146} +42\cdot \frac{42}{146} +44 \cdot \frac{44}{146}\\\\E(X)=\frac{392}{73}+\frac{512}{73}+\frac{882}{73}+\frac{968}{73}\\\\E(X)=\frac{2754}{73} \approx 37.73$

The expected value of Y is

$E(Y)=\sum_{x\in [28,32,42,44]} x\cdot p_{Y}(x)$

$E(Y)=28\cdot \frac{1}{4}+32\cdot \frac{1}{4} +42\cdot \frac{1}{4} +44 \cdot \frac{1}{4}\\\\E(Y)=146\cdot \frac{1}{4}\\\\E(Y)=\frac{73}{2} \approx 36.5$

(b) Let X have probability mass function p(x) and expected value E(X). Then the variance of X, denoted by V(X), is

$V(X)=\sum_{x\in D} (x-\mu)^2\cdot p(x)=E(X^2)-[E(X)]^2$

The variance of X is

$E(X^2)=\sum_{x\in [28,32,42,44]} x^2\cdot p_{X}(x)$

$E(X^2)=28^2\cdot \frac{28}{146}+32^2\cdot \frac{32}{146} +42^2\cdot \frac{42}{146} +44^2 \cdot \frac{44}{146}\\\\E(X^2)=\frac{10976}{73}+\frac{16384}{73}+\frac{37044}{73}+\frac{42592}{73}\\\\E(X^2)=\frac{106996}{73}$

$Var(X)=E(X^2)-(E(X))^2\\\\Var(X)=\frac{106996}{73}-(\frac{2754}{73})^2\\\\Var(X)=\frac{106996}{73}-\frac{7584516}{5329}\\\\Var(X)=\frac{7810708}{5329}-\frac{7584516}{5329}\\\\Var(X)=\frac{226192}{5329} \approx 42.45$